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What is the Cartesian form of #-2+r^2 = -6theta+sin(theta)cos(theta) #?

1 Answer

Shwetank Mauria

May 23, 2016

#x^2+y^2=6tan^(-1)(y/x)+(xy)/(x^2+y^2)+2#

Explanation:

A Cartesian point #(x,y)# in polar form is #(r,theta)#, where

#x=rcostheta# and #y=rsintheta# and hence

#x^2+y^2=r^2cos^2theta+r^2sin^2theta=r^2# and #theta=tan^(-1)(y/x)#

Hence #-2+r^2=-6theta+sinthetacostheta# is

#-2+x^2+y^2=6tan^(-1)(y/x)+(xy)/(x^2+y^2)#

or #x^2+y^2=6tan^(-1)(y/x)+(xy)/(x^2+y^2)+2#

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